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3 years ago

12

A retiree investment $6,000 in a saving plan that pays 4% per year what will the account balance be at the end of the first year

?

1 answer:

allsm [11]3 years ago

3 0

Answer:

$6,240

Step-by-step explanation:

$6,000 x .04 = 240

$6,000 + 240 = $6,240

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What is the work and awnser to 4732 dived by 364

Fiesta28 [93]

The answer is 13. The work is below

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3 years ago

Rachel Newton scored 33 points in a recent basketball game. She scored 21 times, making several free throws worth 1 point each a

SpyIntel [72]

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Step-by-step explanation:

Step 1:

Let Rachel scored 33 points in total

Let a be the 1 point and b be the 2 point.

a = 21 points scored by 1 point. We want to find b

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a+b=33

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3 years ago

Solve y = -7(-13)<br><br> I'm giving 30 points!

VARVARA [1.3K]

y = -7(-13)

=> y = -7 × (-13)

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3 years ago

Read 2 more answers

Paul travel to the lake and back. The trip took 3 hours and the trip back took 4 hours He averaged 10 mph faster on the trip the

ollegr [7]

Answer:

40 mph

Step-by-step explanation:

We assume "outbound" refers to the trip <em>to the lake</em>. The ratio of speeds is inversely proportional to the ratio of times, so ...

outbound speed : inbound speed = 4 : 3

These differ by one ratio unit, so that one ratio unit corresponds to the speed difference of 10 mph. Then the 4 ratio units of outbound speed will correspond to ...

4×10 mph = 40 mph

Paul's average speed on the outbound trip was 40 mph.

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The distance to the lake was 120 mi.

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3 years ago

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Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the

Black_prince [1.1K]

Answer:

a) P(2)=0.270

b) P(X>3)=0.605

c) P=0.410

Step-by-step explanation:

We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the Poisson distribution:

$\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}$

a) In this case: $\lambda=2.1$

$P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270$

Therefore, the probability is P(2)=0.270.

b) In this case: $\lambda=2\cdot 2.1=4.2$

$P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605$

Therefore, the probability is P(X>3)=0.605.

c) We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.

In this case: $\lambda=2\cdot 2.1=4.2$

$P(X\geq 5)=1-P(X$

Therefore, the probability is P=0.410.

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3 years ago